LECTURE OCTOBER, 2016

Transcription

1 LECTURE OCTOBER, 2016 RICHARD MELROSE Abstract. Notes before and after lecture if you have questions, ask! Read: Notes Chapter 2. Unfortunately my proof of the Closed Graph Theorem in lecture was bogus. So as penance I have written out (still fairly brief but now correct I think) proofs of the Three Theorems below. Before lecture Hilbert spaces are separable and infinite dimensional unless otherwise stated! Theorems: Uniform boundedness, Closed graph, Open mapping. I do not plan to prove these, standard proofs rely on Baire s Theorem and can be found in the notes or below. Spectrum of an operator. Neumman series. Group of invertibles, unitary operators, Kuiper s theorem. Again no proof of Kuiper s theorem since we will not use it; see below. Finite rank and compact ideals. Calkin algebra. Hilbert-Schmidt and trace class ideals. Trace functional. Fredholm and semi-fredholm operators. Density. Functional calculus for self-adjoint/normal operators. Spectral theorem for compact self-adjoint operators. Polar decomposition. After lecture Contrary to the statements above I did go quickly through the proofs of Baire s theorem, the Uniform boundedness principle and the Open Mapping Theorem. I may managed to write a quick outline but not yet. I did not talk about unitary operators (yet) nor did I go into Kuiper s Theorem in any detail. I showed that GL(H) B(H) is open (Neumann series). I discussed finite rank operators as a 2-sided ideal in B(H). The subspace K(H) B(H) of compact operators consists of those operators K B(H) such that K(B(0, 1)) M H where M is compact. Proposition 1. K is a closed 2-sided -ideal which is the closure of R(H), the ideal of finite rank operators. Proof. First we show (I did this in class) that R K. Suppose R R n K in norm. By definition the range of R n is finite dimensional so give ɛ > 0 1

2 2 RICHARD MELROSE there exists W n = R n (H) such that (1) Ku R n u < ɛ/2 u B(0, 1). Thus K(B(0, 1)) is a ɛ/2 close to a finite-dimensional subspace, it is bounded and its closure is both bounded and ɛ close to a finite dimensional subspace. Thus K is compact. Conversely suppose K K(H). By definition of compactness, applied to K(B(0, 1)), given ɛ > 0 there is a finite dimensional subspace W such that for each u K(B(0, 1)) there exists w W such that Ku w < ɛ. Let Π W be the orthogonal projection onto W then Ku = Π W Ku + (Id Π W )Ku is the orthogonal decomposition of Ku and Π W Ku W is the element closest to u, that is (2) (Id Π W )Ku < ɛ u B(0, 1) = K Π W K < ɛ. This shows that K R. By the continuity of products and adjoints it follows that K(H) is also an ideal and -closed. I mentioned but did not prove that K(H), {0} and B(H) are the only norm-closed ideals. In fact any other ideal must be contained in K. I did not mention explicitly (but will do on Thursday) that (3) K K = Id K has finite dimensional null space and closed range. An element in the null space of Id K satisfies u = Ku, so the unit ball in the null space is its own image under K. It follows that the unit ball of the null space is contained in a compact set and hence is compact. From the homework this week it is therefore finite dimensional. To see that the range is closed, suppose f n = (Id K)u n f. We must find u H such that (Id K)u = f. Taking the orthogonal decomposition u n = w n + v n with respect to Nul(Id K) (which is closded) with v n Nul(Id K) it follows that f n = (Id K)w n. We proceed to show that w n u from which it follows that (Id K)u = f as claimed. Suppose w m was not bounded, then passing to a subsequence we can arrange that w n. Then setting w n = w n / w n (4) w n Kw n = f m w n 0. Since w n is bounded, Kw n lies in a compact set so has a convergent subsequence and from (4) it follows that, again passing to a subsequence, w n w Nul(Id K). Passing to the limit in (4) however (Id Kw) = 0, so w = 0 which contradicts the fact that w = 1. Thus in fact the sequence w n must be bounded. Applying the same argument but not to w n Nul(Id K) it follows that (5) w n = Kw n + f n. Again Kw n must have a convergent subsequence so w n must have a convergent subsequence with limit u which satisfies (Id K)u = f and we see that the range of Id K is closed. Thus Id K, K K(H) is an example of a Fredholm operator.

3 L11 3 Definition 1. An element P B(H) is Fredholm if it has finite dimensional null space, closed range and P has finite dimensional null space. Recall that for any bounded operator Nul(P ) is the orthocomplement of Ran(P ) whether the latter is closed or not. Certainly P w = 0 implies that (6) (P u, w) = (u, P w) = 0 = Nul(P ) Ran(P ). The same argument can be reversed, that is if (P u, w) = 0 for all u H then P w = 0 which shows that Nul(P ) = (Ran(P )). Note that Nul(P ) is closed but the corresponding orthogonal decomposition is (7) H = Nul(P ) Ran(P ) since Ran(P ) need not be closed in general. In particular one cannot drop the explicit statement that Ran(P ) is closed in the definition of a Fredholm operator you can say Ran(P ) = (Nul(P ) where Nul(P ) is finite dimensional. Proposition 2. An operator is Fredholm if and only if there is a Q B(H) satisfying any one of the following conditions (1) (2) (3) QP = Id K 1, P Q = Id K 2, K 1, K 2 K(H). QP = Id R 1, P Q = Id R 2, R 1, R 2 R(H). (8) QP = Id Π Nul(P ), P Q = Id Π Nul(P ), Π Nul(P ), Π Nul(P ) R. Proof. The last form, (8) implies the preceding one, which in turn implies the first. Moreover the first form implies that P is Fredholm since from the first identity Nul(P ) Nul(QP ) Nul(Id K 1 ) is finite-dimensional from the discussion above. From the second identity Ran(P ) Ran(P Q) = Ran(Id K 2 ) is, again from the discussion above, closed and of finite codimension from which it follows that Ran(P ) is closed of finite codimension. So P is Fredholm. So it suffices to show that if P is Fredholm then there is a bounded operator Q satisfying the (8) (it is in fact unique). We can restrict P to be an operator (9) P : Nul(P ) u P u Ran(P ) where both domain and range now are Hilbert spaces. Clearly P is a bijection and so by the Open Mapping Theorem has a continuous inverse, (10) Q : Ran(P ) Nul(P ), Q P = P Q = Id. Then define Q : H H to be Q(Id Π Nul(P )) which is bounded and satisfies (8). Notice that any operator Q satisfying one of these conditions is also Fredholm.

4 4 RICHARD MELROSE Here is a brief discussion of the three results on operators arising from completeness, so all based on Theorem 1 (Baire). If a non-empty complete metric is written as a countable union of closed subsets (11) M = n C n then one of the C n (at least) must have an interior point. Proof. We find a contradiction to the assumption that none of the C n has an interior point. Start with C 1 M since otherwise it has an interior point so we can find x 1 M \ C, ɛ 1 > 0 s.t. B(x 1, ɛ 1 ) C 1 = where we use the assumption that C 1 is closed, so its complement is open. Next, B(x 1, 1 3 ɛ 1) cannot be contained in C 2 so there must exist (12) x 2 B(x 1, 1 3 ɛ 1), ɛ 2 > 0, ɛ 2 < 1 3 ɛ 1, B(x 2, ɛ 2 ) C 2 =. The conditions ensure that B(x 2, ɛ 2 ) B(x 1, 2 3 ɛ 1) so the smaller ball is disjoint from C 1 as well. Now proceed by induction and so construct a sequence where for all j 2, x j B(x j 1, 1 3 ɛ j 1), ɛ j > 0, ɛ j < 1 3 ɛ j 1, B(x j, ɛ j ) C j =. It follows that ɛ j < 3 1 j ɛ 1 so is summable and hence {x j } is Cauchy so converges by the assumed completeness. The limit x B(x n, ɛ n ) for all n, since the sequence is eventually in the open ball. The conditions above show that x / C n for all n which is the contradiction. Now with this we can prove Theorem 2 (Uniform Boundedness=Banach-Steinhaus). Suppose B and N are respectively a Banach and a normed space and L B(B, N) is a subset of the bounded operators which is pointwise bounded in the sense that (13) for each u B, {Lu; L L} N is bounded then L is bounded (in norm). The fact that N need not be complete is rather bogus since we can always replace it by its completion and nothing changes. Proof. For each n set (14) C n = {u B; u 1, Lu n L L}. The assumption (13) shows that {u B; u 1} = n C n and each C n is closed by the assumtion that each L L is bounded, i.e. continuous. So Baire s Theorem applies and this means there exists ɛ > 0 and u such that (15) v B, v < ɛ = L(u + v) n = Lv n + Lu 2n, L L So in fact L 2n/ɛ for all L L.

5 L11 5 Next in order now is Theorem 3 (Open Mapping). If L : B 1 B 2 is a bounded surjective map between Banach spaces then it is open (16) L(O) B 2 is open for each open set O B 1. Proof. Start with the most important case that L is actually a bijection. Then we try to show that the inverse image of the ball of radius one in B 2 has an interior point in B 1 by setting (17) E N = L({u B 1 ; u < N, Lu < 1}) {f B 2 ; f < 1} = E N by surjectivity. The problem is that we do not know that the E N are closed (this is the same problem that dooms the proof of the Closed Graph Theorem that I tried to give in class, that can presumably be corrected in the same way). So, just let C N be the closure of E N and Baire s Theorem does apply. So for some N there is a ball of positive radius contained in the closure of the image of {u B 1 ; u < N} under L. By surjectivity the centre is the image of some point so subtracting that and scaling using the linearity of L what we conclude is that for some p (18) f B 2, f 1 = u n B 1, u n p, Lu n f. The problem of course is that we do not immediately know that u n u. To see this back off a little from (18) and just use the fact that we can get arbitrarily close with the sequence, then scale again to to see that (19) f B 2 = v B 1, v p f s.t. f Lv 1 2 f. Choose such a v = v 1 and then iteratively choose a sequence f n B 2 v n B 1, where f 0 = f and f n = f n 1 Lv n 1, f n Lv n 1 2 f n, v n p f n = f n f n, n 1. So in fact f n 2 n f and hence v n 2 n p f so the series v n is summable (Cauchy in a complete space) and (20) u = n v n satisfies Lu = n Lv n = n (f n 1 f n ) = f, u 2p f. Since L is bijective this solution is the unique one and this proves that the inverse is bounded and hence the map is open. For the general case of a surjective linear map in the case of Hilbert spaces as domain apply this special result to L : Nul(L) B 2 which is therefore open and check by hand that the projection map Π : H Nul(L) is open. For the general case of a Banach space as domain use quotients instead. Theorem 4 (Closed Graph). A linear map between Banach spaces L : B 1 B 2 is bounded if and only if its graph (21) Gr(L) = {(u, Lu); u B 1 } B 1 B 2 is closed.

6 6 RICHARD MELROSE Proof. If L is bounded, i.e. continuous, then (u n, Lu n ) (u, f) in B 1 B 2 implies u n u and hence Lu n Lu = f so the limit is in Gr(L) which is therefore closed. Conversely suppose the graph is closed, it is then a Banach space with the norm u B1 + Lu B2. The projection operators π 1 and π 2 from B 1 B 2 (which is a Banach space) to B 1 and B 2 are both continuous. By definition of a map the restriction (22) π 1 : Gr(L) B 1 is a bijection and bounded. So when Gr(L) is closed we can use the Open Mapping Theorem to see that (π 1) 1 : B 1 Gr(L) is also bounded. However L = π 2 (π 1) 1 so it is also bounded. Department of Mathematics, Massachusetts Institute of Technology address: rbm@math.mit.edu